COdedorce 366 C Dima and Salad 题解(隐形01背包，好题)

C. Dima and Salad time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Dima, Inna and Seryozha have gathered in a room. That's right, someone's got to go. To cheer Seryozha up and inspire him to have a walk, Inna decided to cook something.

Dima and Seryozha have n fruits in the fridge. Each fruit has two parameters: the taste and the number of calories. Inna decided to make a fruit salad, so she wants to take some fruits from the fridge for it. Inna follows a certain principle as she chooses the fruits: the total taste to the total calories ratio of the chosen fruits must equal k. In other words,  , where aj is the taste of the j-th chosen fruit and bj is its calories.

Inna hasn't chosen the fruits yet, she is thinking: what is the maximum taste of the chosen fruits if she strictly follows her principle? Help Inna solve this culinary problem — now the happiness of a young couple is in your hands!

Inna loves Dima very much so she wants to make the salad from at least one fruit.

Input

The first line of the input contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ 10). The second line of the input contains n integersa1, a2, ..., an (1 ≤ ai ≤ 100) — the fruits' tastes. The third line of the input contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 100) — the fruits' calories. Fruit number i has taste ai and calories bi.

Output

If there is no way Inna can choose the fruits for the salad, print in the single line number -1. Otherwise, print a single integer — the maximum possible sum of the taste values of the chosen fruits.

Examples input 3 2 10 8 1 2 7 1 output 18 input 5 3 4 4 4 4 4 2 2 2 2 2 output -1 Note

In the first test sample we can get the total taste of the fruits equal to 18 if we choose fruit number 1 and fruit number 2, then the total calories will equal 9. The condition  fulfills, that's exactly what Inna wants.

In the second test sample we cannot choose the fruits so as to follow Inna's principle.

题目要求a的和为b的和的k倍，就是说最后要达到的状态为ai1+ai2+ai3+ai4==kbi1+kbi2+kbi3+kbi4……

不妨设ai=k*bi+c，c=k*bi-ai; 这里的c表示如果选了这组，那么还差多少才能满足题意。很显然，最后得到的结果中，所有的c的和应该为0.

定义dp[i]表示还差 i 可以满足题意时 最大能达到的 a （taste值）。

dp[j]=max(dp[j] , dp[j-c[i]]+a[i]);

以  k*bi-ai;作为w ，ai作为v，如果当前有的物品的的sum（k*bi-ai）是0，就说明这个满足a的和为b的和的k倍（移项就好了），ci有正有负，这种题就对正的背包一下，再对负的背包一下（背包的时候 变成正的），然后枚举所有ci（1-10000），如果填满了dp1[]，dp2[]就有值了，直接相加行了，因为他们的ci相同，只是一正一负。。所以相加 = 0

初始化，一开始dp都是-INF， dp[0]是0，这样保证了当前容量一定是填满了，符合要求，最后枚举题目条件（也就是背包的w），如果两个背包填满了，就把两个背包相加，这样就是0了

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int INF = 0x3f3f3f3f; const int maxn = 1e4 + 5; int n, k; int a[maxn], b[maxn], c[maxn], dp1[maxn], dp2[maxn]; int main() { scanf("%d%d", &n, &k); memset(dp1, -0x3f, sizeof(dp1)); //初始化比较重要 memset(dp2, -0x3f, sizeof(dp2)); for(int i = 1;i <= n; i++) scanf("%d", &a[i]); for(int i = 1; i <= n; i++) scanf("%d", &b[i]); for(int i = 1; i <= n; i++) { c[i] = b[i]*k - a[i]; } dp1[0] = dp2[0] = 0; for(int i = 1; i <= n; i++) { if(c[i] >= 0) { for(int v = 10000; v >= c[i]; v--) dp1[v] = max(dp1[v], dp1[v-c[i]]+a[i]); } else { for(int v = 10000; v >= -c[i];v--) dp2[v] = max(dp2[v], dp2[v+c[i]]+a[i]); } } int ans = 0; for(int i = 0; i <= 10000; i++) ans = max(ans, dp1[i]+dp2[i]); if(ans) printf("%d\n", ans); else printf("-1\n"); return 0; }